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Home , Inductance, Magnetomotive force, Permeance

CHAPTER - 2

Learning objectives

lines Force on current carrying conductor Biot Savart's law Permeability and relative permeability Reluctance Comparison of electric circuits and magnetic circuits Composite series magnetic circuits Leakage coefficient Electromagnetic induction Faraday's laws Lenz law Dynamically and statically induced emf Self and mutual Coefficient of in a 45

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Coloumb first determined experimentally the quantitative expression for the magnetic force between two isolated poles. In reality magnetic poles cannot exist in isolation. Thus, the concept is purely theoretical. However, poles of a long thin may be assumed to be isolated poles. The force between two magnetic poles placed in a medium is

(i) directly proportional to their pole strengths m, (ii) inversely proportional to the square of the distance d between them (iii) inversely proportional to the absolute permeability of the medium.

m m Km m F 1 2 or F 1 2 � ∝ µd2 = µd2

1 In SI system of units the value of K is 4π

m1m2 m1m2 F 2 2 N (2.1) � = 4πµd = 4πµ0µr d

where m1, m2 are the pole strengths, d is the distance in m, µ0 is permeability of free 7 space 4π 10 H/m, µr is relative permeability of the medium. = × − Thus, theoretically a unit magnetic pole may be defined as that pole which when placed in vaccum at a distance of one meter from a similar and equal pole repels it with a force of 1 N. discovered in 1820 that a magnetic field is produced around a current 4πµ0 carrying conductor.

2.2.1. Biot-Savart Law The expression for the magnetic field dB produced at a point P by an elemental length dl � � of a conductor carrying a current of I is given by Biot-Savart’s law. Referring to Fig. 2.2.

µIdl sin θ dB Wb/m2 � = 4πr2 or

µIdl ar dB �×� Wb/m2 (2.2) � = 4πr2

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I

dl ar

θ P Figure 2.2 Biot-Savart’s law.

where a is the unit vector along lines joining dl to P . The direction of dB is perpendicular r � to the plane� containing both dl and a . The field at a distance r due to an infinitely long � r straight conductor carrying a current�I amperes is given by

µI B Wb/m2. � = 2πr

The flux lines are in the form of concentric circles around the conductor. If the conductor is held with the thumb pointing in the direction of the current, the encircling fingers give the direction of the magnetic field.

2.2.2. Force on a current carrying conductor It was further observed that another current carrying conductor experiences a force when placed in the field. Now we can recollect that current is nothing but flow of electrons (charges!). Thus magnetic fields are produced by moving charges (current carrying con- ductor) and exert a force on moving charges. The characteristics of this magnetic force on a moving charge are as follows:

Its magnitude is proportional to the magnitude of the charge. • The magnitude of the force is proportional to the magnitude or strength of the field. • The magnetic force depends on the particle’s (charge’s) velocity v. This is different • from the electric field force which is the same whether the charge is� moving or not. A charged particle at rest experiences no magnetic force. By experiment it is found that the force is always perpendicular to both the magnetic • field B and the velocity v. � �

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The above characteristics can be put compactly as,

F qv B (2.3) � = � × � Similarly the force experienced by a current carrying conductor in a magnetic field is found to be proportional to the magnetic field B, the current I and the length of the � conductor and is perpendicular to the field and the length of the conductor. Thus,

F Il B (2.4) � = �× � Since, the direction of the conductor, fixes the direction of the current (in space) (2.4) is more commonly written as

F lI B (2.5) � = � × �

Let F be the force in Newtons, I the current in amperes and l the length of the conductor � � in meters, at right amperes to the magnetic field. Then the magnetic field B or flux density � is the density of a magnetic field such that a conductor carrying a current of 1 at right angles to the field has a force of 1 newton per meter acting upon it. The unit is Tesla (T), after the scientist . The force on a current carrying conductor is given by,

F lIB sin θ (2.6) � = where θ is the angle between the magnetic field and the current carrying conductor. Thus a current carrying conductor experiences a force in the presence of a magnetic field. This principle is used in all electric motors. The direction of the force may be found from Fleming’s left-hand rule as shown in Fig. 2.3.

force F

magnetic field B

current I left hand Figure 2.3

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Hold out your left hand with the fore finger, middle finger and thumb at right angles to each other. If the fore finger represents the direction of the field and the middle finger the direction of the current, the thumb gives the direction of the force on the conductor. From (2.5) it is obvious that no force is exerted on the conductor when it is to the magnetic field (θ 0 ). = ◦

2.2.3. Force between two current carrying conductors

Consider two conductors carrying currents I1 and I2 respectively, separated by a distance of dm. The force between the conductors is attractive if the currents flow in the same direction and repulsive if the currents flow in opposite directions. Let us consider the force on the current due to the first. The field produced by conductor 1 is given by

µI B 1 T = 2πd The force experienced by conductor 2, from (2.5) is given by

µlI I F 2 1 N = 2πd or the force per unit length is given by

µI I F 1 2 N/m. = 2πd

2.2.4. Magnetic flux For a magnetic field having a cross-sectional area Am2 and a uniform flux density of B Teslas, the total flux in Webers (Wb) passing through a plane at right angles to the flow is given by

φ BA = (Webers) (Tesla) (m2) = × or φ B (2.7) = A

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Hence the unit of B is also Wb/m2

1Tesla 1Wb/m2 =

Example 2.1 A conductor carries a current of 500A at right angles to a magnetic field having a density of 0.4T. Calculate the force per unit length on the conductor. What would be the force if the conductor makes an angle of 45◦ to the magnetic field?

Solution: F lI B lIB sin θ = � × � = When conductor is at right angles to the magnetic field, θ 90 . = ◦ F (1m)(500A)(0.4T) 200N/m. = = When θ 45 , = ◦

F (1m)(500A)(0.4T) sin 45◦ = × 141.42N/m. =

Example 2.2 A rectangular coil 100mm by 150mm is mounted so that it rotates about the mid points of the 150mm sides. The axis of rotation is at right angles to a magnetic field with a flux density of 0.02T. Calculate the flux in the coil when (i) Maximum flux links with the coil. What is the position at which this occurs?

(ii) The flux through the coil when the 150mm sides make an angle of 30◦ to the direction of flux.

Solution:

(i) This is shown in Fig. 2.4(a). The maximum flux passes through the coil when the plane of the coil is at right angles to the direction of the flux.

3 3 φ BA 0.02 (100 10− ) (150 10− ) = = × × × × 0.3mWb. =

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axis of rotation

30° 0.02T 0.02T (a) (b) Figure 2.4 Example 2.1.

(ii) This is shown in Fig. 2.4(b).

3 φ BAsin θ (0.3 10− ) sin 30◦ = = × × 0.15mWb. = 2.3. Magnetomotive force and magnetic field strength The magnetic flux is present in a due to the existence of a magnetomotive force (mmf), caused by a current flowing through one or more turns. It is analogous to emf in an electric circuit which is responsible for the .

mmf NI (2.8) = where N is the number of turns. N is a dimensionless quantity. Hence, the unit of mmf is actually Ampere, though more commonly the unit is said to be ampere-turns (AT). Consider a coil as shown in Fig. 2.5. If the magnetic circuit is homogeneous and has a uniform cross sectional area, the mmf per length of the magnetic circuit is called the magnetic field strength H .

NI H AT/m (2.9) = l

The unit of H in SI units is A/m.

The ratio B/H is the permeability, µ0, in free space

B µ0 (2.10) = H

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N turns

Figure 2.5 Coil with N turns on a toroid.

This value is almost same when the conductor is placed in free space, air or in any other non-magnetic material like water, wood, oil etc. 7 µ0 4π 10− H/m (2.11) = × For magnetic materials, B H = µ

µ µ0µr (2.12) = where µr is the relative permeability. The relative permeability is defined as the ratio of the flux density produced in a material to the flux density produced in vacuum by the same magnetic field strength. The relative permeability of non-magnetic materials is close to 1. The relative permeability of magnetic materials is very high, as shown in Table 2.1.

Table 2.1 Relative permeability of magnetic materials.

Material µr Application Ferrite U60 8 UHF chokes Ferrite M33 750 Resonant circuit RM cores Nickel (99% pure) 600 Ferrite N41 3,000 Power circuits Ferrite T38 10,000 Broadband Silicon steel 40,000 Dynamos, mains

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When working with non magnetic materials, the permeability is close to µ0, making it difficult to characterize them by permeability. We make use of magnetic susceptibility defined as

ψm µr 1 (2.13) = −

Example 2.3 A coil of 100 turns is wound uniformly over a wooden ring having a mean circumference of 500mm and a uniform cross sectional area of 500mm2. If the current through the coil is 2.0A calculate (i) the magnetic field strength (ii) the flux density (iii) the flux (iv) mmf

Solution:

(i) Mean circumference 500mm 0.5m. = = NI 100 2 H × 400AT/m or A/m = l = 0.5 =

7 (ii) B µ0H 4π 10 400 502.65 µT = = × − × = (iii)

6 6 φ BA 502.65 10− 500 10− = = × × × 0.2513 µWb = (iv) mmf NI 100 2 200AT. = = × =

Example 2.4 Calculate the mmf required to produce a flux of 0.01Wb across an airgap 2mm long, having an effective area of 100cm2.

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Solution:

2 4 2 A 100cm 100 10− 0.01m = = × = φ 0.01 B 1T = A = 0.01 = B 1 H 7 AT/m = µ0 = 4π 10 × − 3 l 2mm 2 10− m = = × 2 10 3 mmf H l × − 1591.55AT = × = 4π 10 7 = × − 2.4. Reluctance Consider the toroid shown in Fig. 2.5, with a cross-sectional area A m2 and a mean circumference of l , with N turns carrying a current I amperes. We know

φ BA = mmf Hl = φ BA A µr µ0 ∴ mmf = Hl = l or mmf mmf φ = l/µr µ0A = S where l S (2.14) = µ0µr A S is the reluctance of the magnetic circuit and is indicative of the opposition of a magnetic circuit to creation of magnetic flux through it. From (2.14) we can write

mmf φS (2.15) = similar to ’s law. Unit of S is AT/Wb. It is analogous to resistance in electric circuits. The reciprocal of reluctance is called the of the magnetic circuit. Its unit is Wb/AT. It is analogous to conductance in electric circuits.

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2.5. Comparison of magnetic and electric circuits Table 2.2 gives the analogous quantities between electric circuits and magnetic circuits.

Table 2.2 Comparison of magnetic and electric circuits.

Magnetic circuit Electric circuit

1 φ-Flux (Webers) I-Current (Amperes) 2 B-Flux density (Wb/m2) J - (A/m2) l ρl 3 S-Reluctance (At/Wb) R-Resistance () = µ0µr A = A 1 4 P -Permeance (Wb/AT) G-Conductance 1 (mhos) = S = R 5 mmf φS (AT) emf IR () = = 6 Permeability (µ) Conductivity (σ) 1 7 Reluctivity Resistivity (ρ) µ 

There are however some differences between electric circuits and magnetic circuits:

The flux does not flow through the magnetic circuit like the current does in an electric • circuit. In electric circuits if the temperature is maintained a constant, the resistance is a constant • and independent of the current. In a magnetic circuit, the reluctance depends on the flux established through it. The reluctance is small for small values of B and larger for larger values of B. This is because the B–H curve is not a straight line. Flow of electric current requires continuous expenditure of energy but in a magnetic • circuit energy is expanded only in creating the magnetic flux but not in maintaining it. A magnetic circuit stores energy in its field, while an electric circuit dissipates its energy • as heat.

2.6. Composite magnetic circuits We will now discuss composite magnetic circuits; circuits connected in series and parallel.

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la

l1

l2

Figure 2.6 Composite series magnetic circuit.

2.6.1. Composite series circuits Consider the composite circuit of Fig. 2.6, made up of two different sections in series and an air gap. Each section is made of a different material and has its own reluctance. The total reluctance is sum of the individual reluctances

ST S1 S2 Sa = + + l l l 1 2 a (2.16) = µ0µr1A1 + µ0µr2A2 + µ0Aa mmf (2.17) = ST

To find the ampere-turns or the total mmf,

Find the H of each section using • B H (if it is air gap) = µ0 B (for magnetic material) = µ0µr Find the mmf (AT) for each section by • AT Hl = Add these ampere-turns to get the total ampere turns. (Similar to adding emf’s in series!) •

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Leakage flux ring Fringing Flux lines (useful flux)

Air gap Figure 2.8 Leakage flux and fringing.

Solution: The circuit is shown in Fig. 2.9.

3 φ 0.6 10− Wb = × 4 2 A 10 10− m = × φ 0.6 10 3 B × − = A = 10 10 4 × − 0.6Wb/m2 = Cast iron

0.15mm 20cm air gap

Cast steel Figure 2.9 Example 2.5.

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Air gap

B 0.6 5 H 7 4.77 10 AT/m = µ0 = 4π 10 = × × − Total air gap length 0.15 0.15 0.3mm = + = 3 0.3 10− m = × AT required to set up flux in air gap Hl = 5 3 4.77 10 0.3 10− = × × × 143.1AT =

Cast iron section B 0.6 H 7 596.83AT/m = µ0µr = 4π 10 800 = × − × πD π 20 10 2 length of the path πr × × − 0.314m = = 2 = 2 = AT required Hl 596.83 0.314 187.41AT = = × =

Cast steel section B 0.6 H 7 2876.29AT/m = µ0µr = 4π 10 166 = × − × πD length of the path 0.314m = 2 = AT required 2876.29 0.314 903.16AT = × = Total AT required 143.1 187.41 903.16 1233.67AT. = + + =

Example 2.6 A mild steel ring having a cross sectional area of 600mm2 and a mean circumference of 500mm has a coil of 300 turns wound uniformly around it. Calculate (i) the reluctance of the ring (ii) the current required to produce a flux of 800µWb in the ring, if the relative permeability is 400.

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Solution:

(i) l 500 10 3 S − 7 × 6 = µ0µr A = 4π 10 400 600 10 × − × × × − 1.658 106AT/Wb = × (ii) mmf φ S = × 6 6 800 10− 1.658 10 = × × × 1326.4AT = mmf 1326.4 I 4.42A = N = 300 =

Example 2.7 A magnetic circuit comprises three parts in series as follows: (a) A length of 60mm with a cross section area of 50mm2 (b) A length of 30mm with a cross section area of 80mm2 (c) An air gap of length 0.3mm and cross section area of 150mm2. A coil of 2500 turns is wound on part (b) and the flux density in air gap is 0.3T. Assuming that there is no leakage, and the relative permeability µr 1500, estimate the current required in the circuit to produce the flux density. =

Solution:

6 6 φ BcAc 0.3 150 10− 45 10− Wb = = × × = × φl 45 10 6 60 10 3 F mmf φS a − − a a a × 7 × × 6 = = = µ0µr Aa = 4π 10 1500 50 10 × − × × × − 28.6AT = φl 45 10 6 30 10 3 F mmf φS b − − b b b × 7 × × 6 = = = µ0µr Ab = 4π 10 1500 80 10 × − × × × − 8.95AT =

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φl 45 10 6 0.3 10 3 F mmf φS c − − c c c × 7 × × 6 = = = µ0µr Ac = 4π 10 1 150 10 × − × × × − 71.62AT =

F Fa Fb Fc 28.6 8.95 71.62 109.17AT = + + = + + = F 109.17 I 0.04367A 43.67mA. = N = 2500 = = (* Note the large mmf required to set up the flux through air gap as compared to a magnetic material.)

Example 2.8 A wooden ring has a circular cross section of 200 sq mm and a mean diameter of 200mm. It is uniformly wound with 600 turns. If the µr 1, find (1) the field strength produced by a current of 2A (ii) magnetic flux density= (iii) current required to produce a flux density of 0.015Wb/m2.

Solution: (i)

mmf NI 600 2 1200AT = = × = 3 Mean length πd π 200 10− 0.628m = = × × = NI 1200 H 1910.83AT/m = l = 0.628 = (ii)

7 3 2 B µ0µr H 4π 10− 1 1910.83 2.4 10− Wb/m = = × × × = × (iii) The flux density is proportional to the current. A current of 2A produces 0.0024Wb/m2. Therefore current required to produce 0.015Wb/m2 is

2 0.015 × 12.5A. 0.0024 =

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Example 2.9 A in the form of a closed circular ring has a mean length of 20cm and a cross sectional area of 1cm2. The relative permeability of the material is 2200. What current is needed in the coil of 2000 turns wound uniformly around the ring to create a flux of 0.15mWb in the iron? If an air gap of 1mm is cut through the core in a direction perpendicular to the direction of this flux, what current is now needed to maintain the same flux in the air gap?

Solution: Reluctance of core

2 l 20 10− S 7 × 4 = µ0µr A = 4π 10 2200 1 10 × − × × × − 723431.5 AT/Wb = 3 φ 0.15 10− Wb = × 3 mmf φS 0.15 10− 723431.5 108.5AT = = × × = mmf NI = mmf 108.5 I 0.05425A = N = 2000 = 3 1 10− Reluctance of 1mm air gap 7× 4 7957747.1 AT/Wb = 4π 10− 1 1 10− = × 3× × × mmf required to set up flux in air gap 0.15 10− 7957747.1 1193.66AT = × × = Total mmf 108.5 1193.66 = + 1302.16AT = 1302.16 Current required 0.65108A = 2000 =

Example 2.10 An iron ring of mean diameter 20cm, having a cross section area of 3 sq cm is required to produce a flux of 0.45mWb. If µr 1800 find the mmf required. If an air gap of 1mm is made in the ring, how many extra= ampere turns are required to maintain the same flux?

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Solution:

Length of the mean path πd π 0.2 = = × 0.6283m = 3 φ 0.45 10− Wb = × 3 4 2 B φ/A 0.45 10− /3 10− 1.5Wb/m = = × × = B 1.5 H 7 663.154AT/m = µ0µr = 4π 10 1800 = × − × mmf Hl 663.154 0.6283 416.65AT = = × = An air gap of 1mm will need extra mmf.

Bg 1.5 Hg 7 = µ0 = 4π 10 × − 3 1.5 1 10− mmf Hg lg × × 1, 193.66AT = × = 4π 10 7 = × − So additional mmf required when a 1mm air gap is cut is 1,193.66AT.

2.7. Electromagnetic induction—Faraday’s law In 1831, discovered the principle of electromagnetic induction. Two of his experiments are well known. The first experimental setup is shown in Fig. 2.10(a).

S A B G SN G

(a) (b) A Figure 2.10 Electromagnetic induction.

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Magnetic Motion of the conductor field Thumb Forefinger

Induced current

Middle finger Figure 2.11 Fleming’s right hand rule.

Consider a coil of N turns. Let the flux linking the coil change from φ1 Wb to φ2 Wb in t . Now,

Initial flux linkage ψ1 Nφ1 Wb turns = Final flux linkage ψ2 Nφ2 Wb turns = Nφ2 Nφ1 emf induced, e − volts = t N(φ2 φ1) − = t

The above relationship can be written as

dφ dψ e N volts (2.19) = dt = dt

dφ To incorporate Lenz’s law we often write e N dt , meaning the induced emf is set up in a direction such that it opposes the rate of=− change of flux. The emf can be induced in two ways

(i) Dynamically induced emf (or motional emf) (ii) Statically induced emf (or transformer emf)

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LI Average induced emf in the coil is t volts. This current also produces a change in flux, φ from 0 to φWb whose average rate is given by t . From Faraday’s law the induced emf Nφ. Thus, we have = t LI Nφ t = t or

Nφ ψ L (2.21) = I = I

Thus inductance is nothing but the flux linkage per ampere. An alternative expression is

dφ L N = dI

We can also define 1H as the inductance of a coil when a current of 1 ampere through the coil produces a flux linkage of 1Wb turn. Now from (2.15)

mmf NI NI φ = reluctance = S = l/µ0µr A 2 Nφ NI N µ0µr A ∴ L N H (2.22) = I = I(l/µ0µr A) = l

Equation (2.22) gives the expression for L from the geometric parameters of the coil.

2.7.7. Mutually induced emf Consider two coils A and B placed as shown in Fig. 2.12. When the S is closed, some flux produced by A, also links with B. This produces an induced emf in coil B and a current flows through circuit of B, as indicated by the deflection. Since a change of current in one coil is accompanied by a change of flux linked with the other coil inducing an emf in it, it is called mutually induced emf. The two coils are said to have a mutual inductance.

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G S

A B

Figure 2.12 Mutually induced emf.

2.7.8. Mutual inductance The unit of mutual inductance is also . Two coils have a mutual inductance of 1H if an emf of 1volt is induced in one coil when the current through the other coil varies uniformly at the rate of 1A/s. Again consider the two coils of Fig. 2.12.

Let I1 be the current flowing through coil A, which produces a flux φ1. All this flux does not link with coil B. The flux φ11 which links only with coil A is called the leakage flux. The flux φ12 which also links with coil B is called the mutual flux.

φ1 φ11 φ12 (2.23) = + Coefficient of coupling,‘K’ is defined as the ratio of mutual flux to total flux.

φ12 K1 (2.24) = φ1 K 1. The induced emf in coil B is given by ≤ dI1 e2 M . = dt Also

dφ12 e2 N2 = dt dφ12 ∴ M N2 = dI1

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where L1, L2 are the self of coil A and coil B respectively. If K1 K2, then �= in (2.27) we use the geometric mean of K1 and K2;

K K1K2 = 0 K 1. Larger values of coefficient of coupling are obtained with coils which are physically≤ ≤ closer, which are wound or oriented to provide a larger common magnetic flux or which are provided with a common path through a material which serves to concentrate and localize the magnetic flux. Coils with K close to unity are said to be tightly coupled.

Example 2.11 A coil consists of 750 turns. A current of 10A in the coil gives rise to a magnetic flux of 1200µWb. Determine the inductance of the coil and the average induced emf in the coil when the current is reversed in 0.01sec.

Solution:

6 N 750; I 10A; φ 1200 10− Wb = = = × Nφ 750 1200 10 6 L × × − 0.09H = I = 10 = Current reverses from 10A to 10A. − dI 10 ( 10) 20A ∴ = − − = dI 20 e L 0.09 180V. = dt = × 0.01 =

Example 2.12 An air cored has a length of 50cm and diameter of 2cm. Calculate its inductance if it has 1000 turns.

Solution: N 2µ µ A L 0 r = l 2 2 2 πd π (2 10− ) 4 2 A × × 3.14 10− m = 4 = 4 = ×

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2 l 50 10− m = × 2 7 4 (1000) 4π 10− 1 3.14 10− L × × × 2 × × = 50 10− 5 × 78.9 10− H 0.7892mH = × = 2.8. Energy stored in magnetic field In an electric field energy is continuously dissipated. The energy cannot be stored. In a magnetic field on the other hand, we need energy only to set up the initial flux and no energy is required to maintain it. Magnetic field stores the energy which has been used to create the flux. Let the current flowing through a coil of constant inductance L Henrys grow at an I uniform rate from zero to I amperes in t seconds. The average value of current is 2 and I the emf induced in the coil is L t volts. The average power absorbed by the magnetic field is ×  1 LI I Watts 2 × t and the total energy absorbed is 1 LI average power I t × = 2 × t × or 1 W LI 2Joules (2.28) = 2 Now lets consider a more general case where the instantaneous current i increases in a coil having a constant inductance LH. The rate of increase can be uniform or non-uniform. If the current increases by di amperes in dt seconds, the induced emf is given by di e L Volts = dt The energy absorbed is di W i L dt Li di . = dt = ·  

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The total energy absorbed by the magnetic field when the current increases from 0 to I amperes is given by

I 1 1 W Li di L i2 I LI 2 Joules = · = × 2[ ]0 = 2 0 2 A From (2.22) L N µ H . The energy per cubic meter Wf is = l 2 1 2 2 µ 1 2 1 1 B Wf I N µH BH Joules (2.29) = 2 l2 = 2 = 2 = 2 µ

Equation (2.29) can be used only if µr is a constant. Now when the inductive circuit is opened, the current has to reduce to zero and the stored energy released. If there is no in the circuit the energy will be mostly dissipated in the arc across the switch. If there is a resistor, the energy is dissipated as heat in the resistor.

2.9. Dot convention In an , which is a two terminal element, if the current enters the terminal used di as positive reference, the induced L dt is positive, as shown in Fig. 2.13(a). If it enters the coil at the terminal used as negative reference, it is negative, as in Fig. 2.13(b). The dot convention is used to fix the polarity of the voltage induced in a coil due to mutual inductance. A dot is placed on each of the coils. The sign of the voltage due to mutual inductance is as follows:

A current entering the dotted terminal of one coil produces a voltage with a positive reference at the dotted terminal of the second coil.

ii+ −

Ldi v Ldi v v = v = − dt dt − + (a) (b) Figure 2.13 Induced voltage.

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M = 2H i1 i2 + + •

v1 L1 L2 v2

• − − Figure 2.16 Example 2.13

Solution: (i) i2 enters undotted terminal of L2. Hence, mutually induced emf is positive at the undotted terminal of L1. However, v1 is referenced positive at the dotted terminal. Therefore,

di2 v1 M (2)(314)(10 cos 314t) =− dt =− 6280 cos 314tV =− Since i1 0, there is no self induced emf. = (ii) i1 enters the dotted terminal of L1. Hence, v2 is positive at the dotted terminal of L2. However, it is referenced positive at the undotted terminal.

di1 t v2 M (2)( 1)( 8e− ) ∴ =− dt =− − − t 16e− V. =− 2.10. Inductance in series Consider two inductances connected in series as shown in Fig. 2.17(a). Each of the coils has self induced emf and mutually induced emf.

di1 di2 v1 L1 M = dt + dt i1 i2 i = = di v1 (L1 M) ∴ = dt +

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M M L L L L i 1 i 2 i • 1 i 2 •

•• + v1 − + v2 − + v1 − + v2 − (a) v (b) v

Figure 2.17 Inductances in series.

Similarly di v2 (L2 M) = dt + di v v1 v2 (L1 L2 2M) = + = dt + + di Leq = dt Leq L1 L2 2M (2.30) = + + This is called series-aiding connection, where the mutual flux and leakage flux aid each other. In Fig. 2.17(b) M is negative di v1 (L1 M) = dt − di v2 (L2 M) = dt − di v v1 v2 (L1 L2 2M) = + = dt + − Leq L1 L2 2M (2.31) = + − This is called series-opposing connection. The instantaneous energy stored in a coupled circuit is given by

1 2 1 2 W L1i L2i Mi1i2 (2.32) = 2 1 + 2 2 ±

As derived earlier M K√L1L2. =

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Example 2.15 The equivalent of two inductances connected in series is 0.6H or 0.1H, depending on the connection. If L1 0.2H find (i) M (ii) K. =

Solution:

L1 L2 2M 0.6 (i) + + = L1 L2 2M 0.1 (ii) + − = 4M 0.5 or M 0.125H = = L1 0.2H =

From (i) L2 0.6 0.2 0.125 2 0.15H = − − × = 0.125 K M/ L1L2 0.722. = = √0.2 0.15 = ×

Example 2.16 Two identical air cored have 200 turns, length of 25cm and cross section area of 3cm2 each. The mutual inductance between them is 0.5µH. Find the self inductance of each coil and the coefficient of coupling.

Solution: 2 2 7 4 N µ0µr A (200) (4π 10 )(1)3 10 L × − × − 60.318µH = l = 0.25 = L1 L2 L = = 6 M 0.5 10− 3 K × 8.289 10− . = √L1L2 = (60.318 10 6)2 = × × − Example 2.17 A closed iron ring of mean diameter 12cm is made from round iron bar of 2cm diameter. It has a winding of 1000 turns. Calculate the current required to produce a flux density of 1.5Wb/m2 given the relative permeability is 1250. Hence find the self inductance.

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(i)

B 0.5 3 µ 1.2 10− = H = 416.67 = × 3 µ 1.2 10− µr × 7 954.9 = µ0 = 4π 10 = × − (ii)

Nφ π(2 10 2)2 L ; φ B A 0.5 × − = I = × = × 4 4 1.57 10− Wb = × 250 1.57 10 4 L × × − 0.0785H = 0.5 = (iii)

4 φ1 BA 1.57 10− Wb = = × 4 φ2 10% of φ1 0.157 10− Wb = = × 4 dφ φ1 φ2 1.413 10− Wb = − = × dφ 1.413 10 4 e N 250 × − 35.325V = dt = × 0.001 =

Example 2.19 When a voltage of 220V is applied to a coil with a resistance of 50�, the flux linking with the coil is 0.005Wb. If the coil has 1000 turns find the inductance of the coil and the energy stored in the magnetic field.

Solution: V 220 Current 4.4A = R = 50 = Nφ 1000 0.005 L × 1.136H = I = 4.4 = 1 1 Energy stored LI 2 1.136 4.42 11J = 2 = 2 × × =

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Example 2.20 A mild steel ring has a mean diameter of 160mm and a cross section area of 300mm2. Calculate (a) the mmf to produce a flux of 333µWb (b) reluctance (c) relative permeability. The B-H data is given in table below. B(T) 0.9 1.1 1.2 1.3 H(AT/m) 260 450 600 820

Solution: φ 400 10 6 B × − 1.2T = A = 333 10 6 = × − (a) From Table, H 600AT/m = 3 mmf Hl 600 π (160 10− ) = = × × × 301.59AT = (b) mmf φS = mmf 301.59 S 9.057 105AT/Wb = φ = 333 10 6 = × × − (c) B 1.2 3 µ 2 10− = H = 600 = × 3 µ 2 10− µr × 7 1591.5. = µ0 = 4π 10 = × −

Example 2.21 A steel circuit has a uniform cross sectional area of 5cm2 and a length of 25cm. A coil of 120 turns is wound uniformly over it. When the current in the coil is 1.5A, the total flux is 0.3mWb. Find (i) H (ii) µr .

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Solution:

mmf NI 120 1.5 180AT = = × = mmf Hl = NI 180 H 720AT/m ∴ = l = 2.5 10 2 = × − φ 0.3 10 3 B × − 0.6Wb/m2 = A = 5 10 4 = × − B 0.6 B µ0µr H or µr 7 = = µ0H = 4π 10 720 × − × 663.145 =

Example 2.22 A steel ring has a mean circumference of 750mm and a cross sectional area of 500mm2. It is wound with 120 turns (a) Using the table of example 2.20 find the current required to set up a magnetic flux of 630µWb in the ring (b) If the air gap in a magnetic circuit is 1.1mm long and 2000mm2 in cross section, calculate the reluctance of the air gap and the mmf required to send a flux of 700µWb across the air gap.

Solution: (a) φ 630 10 6 B × − 1.26 = A = 500 10 6 = × − From table, using interpolation, H 732AT/m. = Hl 732 750 10 3 I × × − 4.575A = N = 120 = (b) φ 700 10 6 B × − 0.35Wb/m2 = A = 2000 10 6 = × − B 0.35 5 H 7 2.785 10 AT/m = µ0 = 4π 10 = × × −

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(b)

3 N1φ1 12, 000 0.05 10− L1 × × 0.12H = I1 = 5 = 3 N2φ2 15, 000 0.075 10− L2 × × 0.225H = I2 = 5 = (c)

M 0.0675 K 0.411 = √L1L2 = √0.12 0.225 = ×

Example 2.25 Two coupled coils of self inductances 0.6H and 0.16H have a coefficient of coupling 0.8. Find mutual inductance and turns ratio.

Solution:

M K L1L2 0.8√0.6 0.16 = = × 0.248H = N1φ1 I1 = L1 N Kφ N Kφ N L K M 2 1 2 1 2 1 = I1 = N1φ1/L1 = N1 N2 M 0.248 ∴ 0.516 N1 = KL1 = 0.8 0.6 = ×

Questions

(1) Define magnetic flux. (2) What is Biot-Savart’s law? (3) What is the force between two current carrying conductors? (4) Explain mmf and its analogy with emf.

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